Given integer roots of $x^2+mx-n=0$ and $x^2-mx+n=0$. Show $6 \mid n$. Suppose that $m$ and $n$ are integers such that both the quadratic equations $x^2+mx-n=0$ and $x^2-mx+n=0$ have integer roots. Prove that $n$ is divisible by $6$.
I figured out the roots of the equations through quadratic formula-for the first one it is $\frac{-m \pm \sqrt{m^2-4n}}{2}$ and for the second one it was $\frac{m \pm \sqrt{m^2+4n}}{2}$ . 
Then I proved that the discriminant has the same parity as that of $m$ through congruences. Therefore, it was clear that if $m^2-4n$ and $m^2+4n$ were perfect squares,then the roots would be integers. I started with $m^2-4n$ and equated it with $a^2$, where $a$ is some integer. Then I rearranged the terms and factorized things to get $(m-a)(m+a)=4n$. This shows that at least one of them should be even but if one is even then the other should also be even. This shows that $m$ and $a$ are of the same parity. 
I don't know what to do next. I want some help to prove that $n$ is even. I could figure out how to prove that $n$ is divisible by 3 once it is proved that $n$ is even.  
 A: We will be using the following special case of the Vieta Relations: the sum of the roots of $x^2+bx+c=0$ is $-b$, and their product is $c$.
Suppose to the contrary that $n$ is odd. The product of the roots of each equation is therefore odd, so each root is odd. 
The product of the roots in the first equation is $-n$, and in the second it is $n$. So in one equation, not necessarily the first, the product of the roots is congruent to $-1$ modulo $4$, and in the other equation the product of the roots is congruent to $1$ modulo $4$.
In the equation with product of roots congruent to $1$, the roots are both congruent to $1$ modulo $4$, or both are congruent to $-1$. So the sum of the roots is congruent to $2$ modulo $4$.
In the equation with product of roots congruent to $-1$ modulo $4$, the sum of the roots is congruent to $0$ modulo $4$.
So the sums of the roots in the two equations are incongruent modulo $4$.
But in either case, the sum of the roots is $m$ or $-m$. These are both congruent to $2$ modulo $4$, or both congruent to $0$, and we have reached our contradiction.
Remark:  A very similar argument will show divisibility by $3$.
A: Claim 1:
Suppose both $m$ and $n$ were odd, then both $x^2+mx-n=0$ and $x^2-mx+n=0$ cannot have any integer solutions.
Reason:
From the first equation we have $n=x(x+m)$. Thus if $x$ is even, then $n$ must be even and if $x$ is odd then $x+m$ is even, either way $n$ would be even which contradicts the assumption that $n$ is odd. Likewise we can do the second equation. 
Claim 2: $n$ must be even.
Reason:
Let $A$ be the integer root of $x^2+mx-n=0$ and $B$ be the integer root of $x^2-mx+n=0$ . Suppose $n$ was odd. In which case both $A$ and $B$ must be odd (since the product of roots will be $\pm n$). By claim 1, $m$ must be even (otherwise no integer roots).  
Moreover 
\begin{align*}
A^2+mA-n & = 0\\
B^2-mB+n &=0
\end{align*}
This gives us $2n=(A+B)(A-B+m)$. The right hand side is divisible by $4$ so the left hand side must be divisible by $4$ as well. But with $n$ odd that is not possible. 
Thus $n$ must be even.
